$$y(x) = a_0 + a_1 x + a_2 x^2 + ... a_M x_M$$
Textbook states by using the "calculus of the variations" the following is true:
$$\frac{\partial}{\partial y(x)} \int \int (y(x) - t)^2 p(x,t)~ dt~ dx ~~~~~~~~(1)$$
$$=2 \int (y(x)-t)~p(x,t)~ dt = 0~~~~~~(2)$$
How does partial with respect to y(x) remove the integral with respect to dx? What are the detailed steps between (1) and (2)?
I can already understand partial of
$$\frac{\partial}{\partial y(x)}(y(x)-t)^2 = 2(y(x)-t)$$
I just can't figure out how the integral $\int ... dx$ gets removed when the partial is with respect to y(x).
shouldn't it be something like:
$$\frac{\partial}{\partial y(x)} \int \int (y(x) - t)^2 p(x,t)~ dt~ dx$$
$$z=y(x)$$
$$dz=y'(x)dx$$
$$dx = \frac{1}{y'(x)}$$
$$\frac{\partial}{\partial z} \int \int (z - t)^2 p(x,t)~\frac{1}{y'(x)} dt ~dz$$
$$=2 \int (z - t)~ p(x,t)~\frac{1}{y'(x)} dt$$ ???